*It is possible to either move the \(3x\) or the \(4x\). Since it is positive, you would do this by subtracting it from both sides: \(\begin3x 2 &=4x-1\ 3x 2\color &=4x-1\color\ -x 2 & =-1\end\) Now the equation looks like those that were worked before.The next step is to subtract 2 from both sides: \(\begin-x 2\color &= -1\color\-x=-3\end\) Finally, since \(-x= -1x\) (this is always true), divide both sides by \(-1\): \(\begin\dfrac &=\dfrac\ x&=3\end\) You should take a moment and verify that the following is a true statement: \(3(3) 2 = 4(3) – 1\) In the next example, we will need to use the distributive property before solving.*

*It is possible to either move the \(3x\) or the \(4x\). Since it is positive, you would do this by subtracting it from both sides: \(\begin3x 2 &=4x-1\ 3x 2\color &=4x-1\color\ -x 2 & =-1\end\) Now the equation looks like those that were worked before.The next step is to subtract 2 from both sides: \(\begin-x 2\color &= -1\color\-x=-3\end\) Finally, since \(-x= -1x\) (this is always true), divide both sides by \(-1\): \(\begin\dfrac &=\dfrac\ x&=3\end\) You should take a moment and verify that the following is a true statement: \(3(3) 2 = 4(3) – 1\) In the next example, we will need to use the distributive property before solving.*

This isn’t 100% necessary for every problem, but it is a good habit so we will do it for our equations.

In this example, our original equation was \(4x = 8\).

Solve: \(5w 2 = 9\) As above, there are two operations: \(w\) is being multiplied by 5 and then has 2 added to it.

We will undo these by first subtracting 2 from both sides and then dividing by 5.

It is easy to make a mistake here, so make sure that you distribute the number in front of the parentheses to all the terms inside.

Solve: \(3(x 2)-1=x-3(x 1)\) First, distribute the 3 and –3, and collect like terms.To “undo” this, we will add that value to both sides.Solve: \(y-9=21\) This time, 9 is being subtracted from y. \(\beginy-9&=21\ y-9 \color&=21\color\y&=30\end\) Next we will look at what are commonly called “two-step” equations. Wiley Online Library requires cookies for authentication and use of other site features; therefore, cookies must be enabled to browse the site.Detailed information on how Wiley uses cookies can be found in our Privacy Policy.Solving linear equations in algebra is done with multiplication, division, or reciprocals.Using reciprocals, or multiplicative inverse, as well as multiplying and dividing with certain formulas, you can solve linear equation word problems.Linear equations in one variable are equations where the variable has an exponent of 1, which is typically not shown (it is understood).An example would be something like \(12x = x – 5\). By doing this, we will slowly be getting the variable by itself. Solve the equation: \(4x = 8\) In this example, the 4 is multiplying the \(x\).Let’s look at one more two-step example before we jump up in difficulty again.Make sure that you understand each step shown and work through the problem as well.

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